Question: Simplify the following expression: $y = \dfrac{5x^2- 21x+4}{5x - 1}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(5)}{(4)} &=& 20 \\ {a} + {b} &=& &=& {-21} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $20$ and add them together. The factors that add up to ${-21}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${-20}$ $ \begin{eqnarray} {ab} &=& ({-1})({-20}) &=& 20 \\ {a} + {b} &=& {-1} + {-20} &=& -21 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({5}x^2 {-1}x) + ({-20}x +{4}) $ Factor out the common factors: $ x(5x - 1) - 4(5x - 1)$ Now factor out $(5x - 1)$ $ (5x - 1)(x - 4)$ The original expression can therefore be written: $ \dfrac{(5x - 1)(x - 4)}{5x - 1}$ We are dividing by $5x - 1$ , so $5x - 1 \neq 0$ Therefore, $x \neq \frac{1}{5}$ This leaves us with $x - 4; x \neq \frac{1}{5}$.